$ \sum _{n=1}^{\infty }\sin\left(\frac{\left(-1\right)^n}{n}\right) $ Does this sum converge or diverge?

Question

$$ \sum _{n=1}^{\infty }\sin\left(\frac{\left(-1\right)^n}{n}\right) $$ Does this sum converge or diverge?

I tried this:

$$ \sum _{n=1}^{\infty }\sin\left(\frac{\left(-1\right)^n}{n}\right) = \sum _{n=1}^{\infty }\sin\left(\frac{-1}{2n-1}\right) + \sum _{n=1}^{\infty }\sin\left(\frac{1}{2n}\right) $$ and I know that $ \sum _{n=1}^{\infty }\sin\left(\frac{1}{2n}\right)$ diverges. But what about $ \sum _{n=1}^{\infty }\sin\left(\frac{-1}{2n-1}\right)$ ?

I know that if $ \sum a_n $ converges and $ \sum b_n $ diverges then $ \sum \left(a_n+b_n\right) $ diverges.

Answer 1

Hint. Note that $$\sin\left(\frac{\left(-1\right)^n}{n}\right)=(-1)^n\sin\left(\frac{1}{n}\right)$$ and the sequence$\{\sin(1/n)\}_{n\geq 1}$ is positive and decreasing. Then apply Leibniz criterion.

P.S. About the odd subsequence: $\sum _{n=1}^{\infty }\sin\left(\frac{-1}{2n-1}\right)=-\sum _{n=1}^{\infty }\sin\left(\frac{1}{2n-1}\right)=-\infty$.